At 0+150 GL = 109.55 and bed = 109.72. What type of section is it?

Full banking, since e = -0.17 m

Full cutting since |e| is below D

Neither, since e is not a whole number

How do the unequal top widths of the banks (3 m left, 1.5 m right) enter the bank area formula?

Through the sum t_L + t_R = 4.5 m multiplied by bank height

Only the larger width, 3 m, is used

The average 2.25 m is used for both banks

They are ignored because slopes decide the area

In a full banking section, the width of permanent land is measured as

The distance between the outer toes of the two banks

The top width of the banks only

Asked 2/22
2080 Baisakh · 12 marks

Channel earthwork and land area, bed 4 m

Short answer

Earthwork: cutting = 27.77 m³, banking = 2,316.23 m³. Permanent land obtained from mean width × length = 2,330.70 m².

Calculate the quantity of earthwork and area of permanent land required for the land acquisition purpose for a portion of a channel from following data: Bed width = 4 m; Free board = 45 cm; Side slope in cutting = 1:1; Side slope in banking = 1.5:1; Full supply depth = 1 m; Top width of bank = 3 m left and 1.5 m right.

Chainage	90	120	150	180	210	220
RL Ground	109.8	109.7	109.55	109.30	109.15	109.10
RL Bed	109.52	Bed slope 1:300 rising

Given data and assumptions

Bed width $4$ m; FSD $1$ m; free board $0.45$ m, so $D = 1.45$ m; top width of bank $3$ m (left) and $1.5$ m (right).
Side slope in cutting $1:1$ , in banking $1.5:1$ (inner and outer faces of banks).
Bed level 109.52 m at chainage 90, rising at 1 in 300 with chainage (assumed direction): 109.52, 109.62, 109.72, 109.82, 109.92, 109.953 m.

Method

Total depth of canal section $D = \text{FSD} + \text{free board} = 1.00 + 0.45 = 1.45$ m. At each station the bed level is compared with the ground level; $e =$ GL − bed level.

If $e \ge D$ : full cutting, area $A_c = (b + s_c e)\,e$ (no banks needed).
If $0 < e < D$ : part cutting, part banking. Cutting area $A_c = (b + s_c e)e$ ; each bank stands $h_b = D - e$ above ground, so bank area (both banks) $A_b = (t_L + t_R)h_b + (s_i + s_o)h_b^2$ with $s_i, s_o$ the inner and outer bank slopes.
If $e \le 0$ : full banking, bed above ground. Bank top is $H = D - e$ above ground. Gross embankment $= (T + s_o H)H$ with $T = t_L + t_R + b + 2 s_i D$ , and the canal space $(b + s_i D)D$ is deducted.

Volume between stations = mean area × distance (separately for cutting and banking).

Sectional areas

Station	GL (m)	Bed RL (m)	$e$ (m)	Condition	Cutting area (m²)	Bank area (m²)
0+090	109.800	109.520	0.280	part cutting, part banking	1.198	9.372
0+120	109.700	109.620	0.080	part cutting, part banking	0.326	11.796
0+150	109.550	109.720	-0.170	full banking	0.000	15.800
0+180	109.300	109.820	-0.520	full banking	0.000	22.182
0+210	109.150	109.920	-0.770	full banking	0.000	26.966
0+220	109.100	109.953	-0.853	full banking	0.000	28.602

Volumes

Between	Distance (m)	Cutting (m³)	Banking (m³)
0+090 – 0+120	30.0	22.87	317.51
0+120 – 0+150	30.0	4.90	413.93
0+150 – 0+180	30.0	0.00	569.73
0+180 – 0+210	30.0	0.00	737.22
0+210 – 0+220	10.0	0.00	277.84
Total		27.77	2,316.23

Permanent land

Width of land at a section = distance between the outer toes of the banks (banking) or canal top width plus the two bank/berm widths (cutting).

Station	Land width (m)
0+090	16.080
0+120	16.880
0+150	17.710
0+180	18.760
0+210	19.510
0+220	19.760

Area of land $=$ mean width × length, summed over the reaches $= 2,330.70$ m² $= 0.2331$ ha.

Answer: Earthwork in cutting = 27.77 m³; earthwork in banking = 2,316.23 m³; total = 2,344.00 m³. Permanent land = 2,330.70 m².

✓ Answer checked by Claude Opus 5.5 (high effort)

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